华为的FPGA和Verilog的教程，我相信对大家的用处是毋庸置疑的。...

分类中没有! 是Fortran 的源码,来源于国外!-classification, no! Yes Fortran source code, from abroad!

mc51单片机开发 8051单片机c语言控制与应用-mc51 Singlechip 8051 c language development and application of control

适合新手学习FPGA开发，针对altera公司的产品。-Suitable for novices to learn FPGA development, for altera company

ChipScope教程！~本人觉得还可以-ChipScope Tutorial! ~ I think that can also be

安装JBuilder以后注册与调整光标的片子,解决很重要的小问题!-JBuilder after installation and adjustment of the cursor registration of the film is very important to solve the small problems!

windows 内核介绍。-windows kernel introduced.

在Windows下的各种网络编程接口中，Windows Sockets脱颖而出，越来越得到大家的重视，这是因为Windows Sockets规范是一套开放的、支持多种协议的Windows下的网络编程接口。从1991年的1.0版到1995年的2.0.8版，经过不断完善并在Intel、Microsoft、Sun、SGI、Informix、Novell等公司的全力支持下，已成为Windows网络编程的事实上的标准。 在作者利用Windows Sockets规范进行应用开发的过程中，发现这方面的资料很少，特别是缺乏一本全面而实用的专著。为了使广大用户能够充分理解和应用这套规范，我们编写了这本书。本书不但对Windows Sockets 1.1及2.0规范作了较为详尽的介绍，还结合了作者的实际工作，给出了具有实际应用价值的程序实例-In the author standard carries on the application development using Windows the Sockets in the process, discovered that this aspect the material are very few, specially lacks one comprehensive and the practical monograph. In order to enable the user community to understand and to apply this set of standards fully, we have compiled this book. Not only this book 1.1 and 2.0 standards has made a more exhaustive introduction to Windows Sockets, but also unified author s practical work, gave had the practical application value procedure example. Hoped that it can standard and applies the function which to Windows the Sockets in

java书 java书 java书 java书-java book book java java book book java book j ava book java java book on java book

这是一本关于C语言的书籍，希望能够帮助更多的人得到进步与提升。-This is a book on the C language, I hope to help more people to be progress and upgrade.

背包问题动态规划算法JAVA 给定n种物品和一背包。物品i的重量是wi，其价值为vi，背包的容量为c。问应如何选择装入背包中的物品，使得装入背包中物品的总价值最大？ 在选择装入背包的物品时，对每种物品i只有两种选择，即装入背包或不装入背包。不能将物品i装入背包多次，也不能只装入部分的物品i。因此，该问题称为0-1背包问题。-Knapsack problem dynamic programming algorithm for a given n species JAVA goods and one backpack. Item i is the weight of wi, the value for vi, knapsack capacity c. Question should be how to choose the items into the backpack, the backpack into the total value of the largest items? In the selection of the items into the backpack, for each of the items i have only two choices, that is loaded or not loaded backpack backpack. Items should not be many times i load backpack can not only part of the goods loaded i. Therefore, the problem known as the 0-1 knapsack problem.